ExplanationThe probability of independent events A and B occurring is equal to the product of the probability of event A and the probability of event B.
In this case, the probability of the coin turning up heads is \(\frac{1}{2}\) and the probability of rolling a 6 is \(\frac{1}{6}\). Therefore, the probability of heads and a 6 is equal to \(\frac{1}{2}\times\frac{1}{6}=\frac{1}{12}\).
Alternatively, list all the possible outcomes: H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6.
There are 12 total outcomes and only 1 with heads and a 6. Therefore, the desired outcome divided by the total number of outcomes is equal to \(\frac{1}{12}\).
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